Linked List/Queue/Stack Pending –

1. circular

2. round petrol problem

3. linked list and stack and queue mixed problem

4. cracking the coding interview problem

5. InterviewBit Problem Set

Make it a point to maintain a tail node in DLL and a size variable.

- ** Doubly Linked List

- **Reverse Doubly Linked List Code

- *** Sort a linked list of 0,1 and 2’s – count 0, 1 and 2’s.

**Algo:**

– We can count the number of nodes having 0,1,2 and then make a new list with equal number of nodes.

– We can do it on O(n) with no extra space. Code. The way this code works is without interchanging data and linking up nodes. It creates there node, one each for 0,1 and 2. Then start traversing the list. Depending on the data of the node we link the current node to the specific node – either 0/1/2. This logic can be used for singly linked list as well as we don’t rely on traversing from in the opposite direction.

– The third logic is same as used for sorting an array of 0,1 and 2. Since we can traverse backwards in the DDL, we will use that to our use.

Dutch national flag sorting problem/ Sort an array of 0s, 1s and 2s.

(1) Create a**low**pointer at the beginning of the array and a**high**pointer at the end of the array.

(2) Create a**mid**pointer that starts at the beginning of the array and iterates through each element.

(3) If the element at arr[mid] is a 2, then swap arr[mid] and arr[high] and decrease the**high**pointer by 1.

(4) If the element at arr[mid] is a 0, then swap arr[mid] and arr[low] and increase the**low**and**mid**pointers by 1.

(5) If the element at arr[mid] is a 1, don’t swap anything and just increase the**mid**pointer by 1.

- ** Swap Kth node from beginning with Kth node from end in a Singly Linked List –

GFG**. Algo –**

Let X be the kth node from beginning and Y be the kth node from end.

Following are the interesting cases that must be handled.

**1)**Y is next to X

**2)**X is next to Y

**3)**X and Y are same

**4)**X and Y don’t exist (k is more than number of nodes in linked list)

- ***
**Design a stack with operations on middle element.**

How to implement a stack which will support following operations in O(1) time complexity? Link

1) push() which adds an element to the top of stack.

2) pop() which removes an element from top of stack.

3) findMiddle() which will return middle element of the stack.

4) deleteMiddle() which will delete the middle element.

Implementing a stack using a Singly Linked List is pretty straightforward. So what we do is to maintain a middlePointer(ok, that’s obvious) and have a flag which indicates when to move the middle pointer. We move the middle pointer when there was an element addition and the total node count became odd.

When the list is empty middlePointer points to the first element. Now when the 3 node is added the middlePointer should be moved to it’s next.

When an element is poped we may have to readjust the middlePointer. If there was even number of element, after pop it becomes odd, so now the middle needs to be moved back by one node. Since we need to move back to maintain the middlePointer in O(1) we will use a double linked list.

C Code here is pretty clean

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